Question

Water's heat of fusion is 80. cal/g

c
a
l
/
g
, its specific heat is 1.0 calg⋅∘C
c
a
l
g
⋅
∘
C

, and its heat of vaporization is 540 cal/g
c
a
l
/
g
.
A canister is filled with 330 g
g
of ice and 100. g
g
of liquid water, both at 0 ∘C
∘
C
. The canister is placed in an oven until all the H2O
H
2
O
has boiled off and the canister is empty. How much energy in calories was absorbed?

Answer 1

The total energy absorbed to heat 330 g of ice and 100 g of liquid water from 0 °C until it has all boiled off is 80400 calories.

Step-by-step explanation:

To calculate the energy absorbed when 330 g of ice and 100 g of liquid water at 0 °C are heated until all the H2O has boiled off, we need to consider the heat of fusion, specific heat, and heat of vaporization of water. The energy to melt the ice is found by multiplying the mass of ice by the heat of fusion. For the boiling water, we use the heat of vaporization. The heat absorbed by the liquid water as it heats from 0 °C to 100 °C is assumed to be zero because it starts at 0 °C and needs to reach 100 °C to begin the vaporization process.

The energy to melt the ice is 330 g × 80 cal/g = 26400 cal. For the water to vaporize, the energy needed is 100 g × 540 cal/g = 54000 cal. The total energy absorbed is therefore 26400 cal + 54000 cal = 80400 cal.